Python Matching Complex String with Regex

Question

I have the following test string:

test_str = `It isn't directed at all,' said the White Rabbit;

My current regular expression uses re.sub to filter out the punctuation so that I can do my own operations.

My current regex is re.sub(r"[^A-Za-z0-9'\s]", '', test_str)

The output from above is:

['It', "isn't", 'directed', 'at', "all'", 'said', 'the', 'White', 'Rabbit']

The error can be seen at all' when it is suppose to be storing all only.

How do you store words with 's and also ignore ' that comes after a punctuation? In this case, all,'.

An0n1m1ty · Accepted Answer · 2020-01-24 01:46:22Z

1

Try the following:

import re
test_str = "`It isn't directed at all,' said the White Rabbit;"
a = re.sub(r"[^A-Za-z0-9'\s]", '', test_str)
a = re.sub(r"'[ ]", ' ', a)
print(a)

answered Jan 24, 2020 at 1:46

An0n1m1ty

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U13-Forward · Accepted Answer · 2020-01-24 01:48:52Z

1

Try using this regular expression:

print(re.sub('["!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~''](?!\w+)', '', test_str))

Output:

It isn't directed at all said the White Rabbit

answered Jan 24, 2020 at 1:48

U13-Forward

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1 Comment

z.yea Over a year ago

I tried working on it but its a little bit complex from what i wanna achieve. Thank you so help for ur help !

Dejoskii · Accepted Answer · 2020-05-05 13:04:23Z

0

Here are other solutions

re.sub("\'[^\w]",' ', test_str)
re.sub("\'[\s]",' ', test_str)
re.sub("\'(?!\w)",'', test_str)
re.sub("\'(?=\s)",'', test_str)

edited May 5, 2020 at 13:04

answered May 5, 2020 at 11:32

Dejoskii

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